$ \left(\dfrac{125}{27}\right)^{-\frac{4}{3}}$
Solution: $= \left(\dfrac{27}{125}\right)^{\frac{4}{3}}$ $= \left(\left(\dfrac{27}{125}\right)^{\frac{1}{3}}\right)^{4}$ To simplify $\left(\dfrac{27}{125}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left(? \right)^{3}=\dfrac{27}{125}$ To simplify $\left(\dfrac{27}{125}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left({\dfrac{3}{5}}\right)^{3}=\dfrac{27}{125}$ so $ \left(\dfrac{27}{125}\right)^{\frac{1}{3}}=\dfrac{3}{5}$ So $\left(\dfrac{27}{125}\right)^{\frac{4}{3}}=\left(\left(\dfrac{27}{125}\right)^{\frac{1}{3}}\right)^{4}=\left(\dfrac{3}{5}\right)^{4}$ $= \left(\dfrac{3}{5}\right)\cdot\left(\dfrac{3}{5}\right)\cdot \left(\dfrac{3}{5}\right)\cdot \left(\dfrac{3}{5}\right)$ $= \dfrac{9}{25}\cdot\left(\dfrac{3}{5}\right)\cdot \left(\dfrac{3}{5}\right)$ $= \dfrac{27}{125}\cdot\left(\dfrac{3}{5}\right)$ $= \dfrac{81}{625}$